Question

The two equal sides of an isosceles triangle where it meets, from that vertices a perpendicular is drawn to the opposite side with a length of 8m. The perimeter of triangle is 64m. What is its area?​

Answer 1

AnswEr :

⋆ Reference of Image is in the Attachment :

• Height ( h ) = 8 m
• Perimeter = 64 m
• Area of Triangle ?

• According to the Question Now :

⇝ Perimeter = AB + BC + AC

⇝ Perimeter = a + b + a

⇝ 64 = 2a + b

⇝ b = 64 - 2a ⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq.( I )

• In ∆ ADC, using Pythagoras Theorem :

⇒ AC² = AD² + CD²

⇒ ( a )² = ( 8 )² + ( b /2)²

⇒ a² = 64 + b² /4

⇒ a² = ( 256 + b² ) /4

⇒ 4a² = 256 + b²

• Putting the value of b from eq.( I )

⇒ 4a² = 256 + ( 64 - 2a )²

⇒ 4a² = 256 + { (64)² + (2a)² - 2 × 64 × 2a }

⇒ 4a² = 256 + 4096 + 4a² - 256a

⇒ 4a² - 4a² + 256a = 256 + 4096

⇒ 256a = 4352

• Dividing Both term by 256

⇒ a = 17 m

• Putting the value of a in eq.( I ) :

⇝ b = 64 - 2a

⇝ b = 64 - (2 × 17)

⇝ b = 64 - 34

⇝ b = 30 m

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• Now Area of the Isoceles Triangle :

⇒ Area = 1 /2 × Base × Height

⇒ Area = 1 /2 × b × h

⇒ Area = 1 /2 × 30 m × 8 m

⇒ Area = 15 m × 8 m

⇒ Area = 120 m²

∴ Area of Isosceles Triangle is 120m².

Answer 2

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Given:

The two equals sides of an isosceles triangle where it meets, from that vertices a perpendicular is drawn to the opposite side with a length of 8m,

The perimeter of triangle is 64m.

To find:

The area of an Isosceles triangle.

Explanation:

We have,

• Perimeter of triangle= 64m
• The length of perpendicular of Isosceles Δ.

We know that formula of the perimeter of Isosceles Δ:

→ Perimeter = 2a + b

→ 64m = 2a + b

→ b = 64 - 2a.......................(1)

Two equal sides of an Isosceles triangle:

Using Pythagoras Theorem:

[Hypotenuse]² = [Base] + [Perpendicular]

• In one part, right angled ΔADC, we get;
• Hypotenuse= AC
• Base= CD= b/2
• Perpendicular= AD=8m

→ AC² = CD² + AD²

→ a² = (b/2)² + 8²

Putting the value of b in this base, we get;

→ a² = $(\frac{64-2a}{2} )^{2}$ + 64

• [(a-b)² = a² + b² -2ab]

→ a² = $\frac{(64)^{2}+(2a)^{2} -2.64.2a }{4}$ +64

→ 4a² = 4096 + 4a² -256a + 256

→ 4a² -4a² + 256a = 4096 +256

→ 0 +256a = 4352

→ 256a = 4352

→ a = $\cancel{\frac{4352}{256} }$

→ a = 17m.

Therefore,

Putting the value of a in equation (1),we get;

→ b = 64m - 2(17m)

→ b = 64m - 34m

→ b = 30m

We obtained the two value of isosceles Δ;

• Base of Isosceles Δ = 30m
• Side of Isosceles Δ = 17m

Now,

We know that area of Isosceles Δ: base × height/2

⇒ Area of Isosceles Δ = $(\frac{30*8}{2}) m^{2}$

⇒ Area of Isosceles Δ = $(\frac{\cancel{30}*8}{\cancel{2}} )m^{2}$

⇒ Area of Isosceles Δ = (15 × 8)m²

⇒ Area of Isosceles Δ = 120m².